I am wondering if my proof is correct? Thank you for whoever willing to take a look at it for me.
Proof $\liminf E_k \subset \limsup E_k $
If $\{E_k\}_{k=1}^\infty$ is a sequence of sets, we define
\begin{align*}
\limsup E_k & = \bigcap_{j=1}^\infty\left(\bigcup_{k=j}^\infty E_k\right)\\
&= \bigcap_{j=1}^\infty(E_j \cup E_{j+1} \cup \cdots)\\
&= (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots.
\end{align*}
Therefore, $\limsup E_k$ consists of those points in $\mathbb{R}^n$ which belong to infinitely many $E_k$.
\begin{align*}
\liminf E_k & = \bigcup_{j=1}^\infty\left(\bigcap_{k=j}^\infty E_k\right)\\
&= \bigcup_{j=1}^\infty(E_j \cap E_{j+1} \cap \cdots)\\
&= (E_1 \cap E_2 \cap \cdots) \cup (E_2 \cap E_3 \cap \cdots) \cup \cdots.
\end{align*}
Therefore, $\liminf E_k$ consists of those points in $\mathbb{R}^n$ which belong to all $E_k$ for $k \geq k_0$.
Now consider $x \in \liminf E_k$, then $x \in E_1 \cap E_2 \cap \cdots$, or $x \in E_2 \cap E_3 \cap \cdots$ and so on, that is to say, $x \in E_{k_0} \cup E_{k_0+1} \cup \dots$ for some $k_0$. More specifically, $x \in E_k \forall k \geq k_0$. Therefore, $x \in (E_1 \cup E_2 \cup \cdots) \cap (E_2 \cup E_3 \cup \cdots) \cap \cdots$, so we showed $x \in \limsup E_k$.
Best Answer
It’s not quite right: you’ve actually assumed that $x\in\limsup_kE_k$, so the argument is circular. Recall that
$$\liminf_{k\in\Bbb N}E_k=\bigcup_{n\ge 0}\bigcap_{k\ge n}E_k\;;$$
this means that if $x\in\liminf_{k\in\Bbb N}E_k$, then there is some $n_0\in\Bbb N$ such that $x\in\bigcap\limits_{k\ge n_0}E_k$.
Since $$\limsup_{k\in\Bbb N}E_k=\bigcap_{n\ge 0}\bigcup_{k\ge n}E_k\;,$$ you have to use this somehow to show that $x\in\bigcup\limits_{k\ge n}E_k$ for every $n\in\Bbb N$. That’s not actually very hard: we know that $x\in E_k$ for each $k\ge n_0$, so if we just take $k=\max\{n,n_0\}$ then ... ?